Hi I have been reading the latest practical everyday electronics (epemag) on the lateset power amplifier the SC200 which was deveoped by the australian magazine siliconchip who gave permission for epemag to use the same design article. I posted a couple of questions which but havent had a reply back.Can anyone else help with the questions below. I have also posted the circuit diagram.
1.) does it matter which version of BC546 is used, ie BC546B, BC546C and also can BC556B be used for the SC200 power amplifier.
2.) i have a query about the 4.4mV drop each emitor resistor Re. I read
elsewhere specifically in Randy slone power amplifier construction
manual and douglas self power amplifier handbook that the optimal V drop
for low distortion with emitter follower output stages need to have a
much higher voltage drop across both emiter resistor ie Randy slone
quotes a value of 47mV and Douglas self quotes 50mV. With Re restors of
0.1 ohms this means the current per transistor pair should be 0.047 /
(0.1 x 2) = 0.235 Amps. So total current for both sets of ouput
transistors = 2 x 0.235 = 0.47 Amps. These values do seem ridiculously
high to me but the fact that 2 different Authors have quoted very
similar Re Voltage drops for emiter follower outputs circuits raises the
question why this design has only 4.4mV.Am i getting confused misunderstood about emiter follower output stages and the optimal biasing required ????.
RegardsHeath
1.) does it matter which version of BC546 is used, ie BC546B, BC546C and also can BC556B be used for the SC200 power amplifier.
2.) i have a query about the 4.4mV drop each emitor resistor Re. I read
elsewhere specifically in Randy slone power amplifier construction
manual and douglas self power amplifier handbook that the optimal V drop
for low distortion with emitter follower output stages need to have a
much higher voltage drop across both emiter resistor ie Randy slone
quotes a value of 47mV and Douglas self quotes 50mV. With Re restors of
0.1 ohms this means the current per transistor pair should be 0.047 /
(0.1 x 2) = 0.235 Amps. So total current for both sets of ouput
transistors = 2 x 0.235 = 0.47 Amps. These values do seem ridiculously
high to me but the fact that 2 different Authors have quoted very
similar Re Voltage drops for emiter follower outputs circuits raises the
question why this design has only 4.4mV.Am i getting confused misunderstood about emiter follower output stages and the optimal biasing required ????.
RegardsHeath